Question prove the following identity showing all steps:\frac{cos(x+30)-sin(x+60)}{sin(x)cos(x)} =-sec(x) please help fast

Answer: See solution below Step-by-step explanation: Given the expression [tex]\frac{cos(x+30)-sin(x+60)}{sin(x)cos(x)} \\[/tex] Recall that cos x = sin(90-x) cos(x+30 ) = sin (90-(x+30) = sin(90-x-30) = sin(60-x) Substitute [tex]\frac{sin(60-x)-sin(x+60)}{sin(x)cos(x)} \\= \frac{sin60cosx-cos60sinx)-sinxcos60-cosxsin60)}{sin(x)cos(x)} \\= \frac{-2cos60sinx)}{sin(x)cos(x)} \\= \frac{-2(1/2)sinx)}{sin(x)cos(x)} \\= \frac{-1}{cos(x)}\\= \frac{1}{cos(x)}\\ \\= -sec(x) Proved[/tex] Log in to Reply

Answer:See solution below

Step-by-step explanation:Given the expression

[tex]\frac{cos(x+30)-sin(x+60)}{sin(x)cos(x)} \\[/tex]

Recall that

cos x = sin(90-x)

cos(x+30 ) = sin (90-(x+30)

= sin(90-x-30)

= sin(60-x)

Substitute

[tex]\frac{sin(60-x)-sin(x+60)}{sin(x)cos(x)} \\= \frac{sin60cosx-cos60sinx)-sinxcos60-cosxsin60)}{sin(x)cos(x)} \\= \frac{-2cos60sinx)}{sin(x)cos(x)} \\= \frac{-2(1/2)sinx)}{sin(x)cos(x)} \\= \frac{-1}{cos(x)}\\= \frac{1}{cos(x)}\\ \\= -sec(x) Proved[/tex]